# Thread: integral help

1. ## integral help

Hello, I am to do the following integral two different wants. Now one way that I picked, which was the easiest was product to sum trig identities. I found the correct answer. Now, Im trying to use integration by parts, and the general idea, is integrate until you get back to your same integral, and then algebraically solve but I can't seem to get there. Here is the problem:

$\int sin(4x)cos(5x)dx$

Now, I get to a point(and please correct me if I'm wrong) where I'm at

$\int sin(4x)cos(5x)dx = {\frac{1}_{5}}sin(4x)sin(5x)-{\frac{4}_{20}}sin(4x)sin(5x)+ \int sin(4x)cos(5x)dx$

But from there.. clearly I can't do anything, if I subtract to the other side, that would defeat my purpose by getting rid of the integral. Maybe I'm doing it completely wrong? Help would be much appreciated!

2. Originally Posted by intervade
Hello, I am to do the following integral two different wants. Now one way that I picked, which was the easiest was product to sum trig identities. I found the correct answer. Now, Im trying to use integration by parts, and the general idea, is integrate until you get back to your same integral, and then algebraically solve but I can't seem to get there. Here is the problem:

$\int sin(4x)cos(5x)dx$

Now, I get to a point(and please correct me if I'm wrong) where I'm at

$\int sin(4x)cos(5x)dx = {\frac{1}_{5}}sin(4x)sin(5x)-{\frac{4}_{20}}sin(4x)sin(5x)+ \int sin(4x)cos(5x)dx$

But from there.. clearly I can't do anything, if I subtract to the other side, that would defeat my purpose by getting rid of the integral. Maybe I'm doing it completely wrong? Help would be much appreciated!
Please post your working. You've probably dropped a minus somewhere.

3. Alright, so for

$\int sin(4x)cos(5x)dx$

using

$uv - \int vdu$

u = sin(4x)
dv = cos(5x)dx
du = 4cos(4x)dx
v = (1/5)sin(5x)

$\int sin(4x)cos(5x)dx = {\frac{1}_{5}}sin(4x)sin(5x) - {\frac{4}_{5}}\int sin(5x)cos(4x)dx$

again..
u = sin(5x)
dv = cos(4x)dx
du = 5cos(5x)dx
v = (1/4)sin(4x)

$\int sin(4x)cos(5x)dx = {\frac{1}_{5}}sin(4x)sin(5x) - {\frac{4}_{5}}[{{\frac{1}_{4}}sin(4x)sin(5x)}-{\frac{5}_{4}}\int sin(4x)cos(5x)dx]$

4. Originally Posted by intervade
Alright, so for

$\int sin(4x)cos(5x)dx$

using

$uv - \int vdu$

u = sin(4x)
dv = cos(5x)dx
du = 4cos(4x)dx
v = (1/5)sin(5x)

$\int sin(4x)cos(5x)dx = {\frac{1}_{5}}sin(4x)sin(5x) - {\frac{4}_{5}}\int sin(5x)cos(4x)dx$

again..
u = sin(5x)
dv = cos(4x)dx
du = 5cos(5x)dx
v = (1/4)sin(4x)

$\int sin(4x)cos(5x)dx = {\frac{1}_{5}}sin(4x)sin(5x) - {\frac{4}_{5}}[{{\frac{1}_{4}}sin(4x)sin(5x)}-{\frac{5}_{4}}\int sin(4x)cos(5x)dx]$
Add $\frac{5}{4} \int \sin(4x) \cos(5x)$ to both sides

5. Hmm, don't I need to distribute the -(4/5)?

6. Originally Posted by intervade
Hmm, don't I need to distribute the -(4/5)?
Oops you would, I missed that set of brackets >.<